7. Multiple integral

Riemann sum

Very similar to this: Riemann Sum (univariate calculus).

Intuition define:

Now give you a partition $P$:

$$P: I=\bigcup_{j=1}^{m} I_{j}$$

of this interval with distinguished points $\xi=\left\{\xi_{1}, \cdots, \xi_{k}\right\}$.

We defined Riemann sum as:

$$\sigma(f, P, \xi):=\sum_{i=1}^{k} f\left(\xi_{i}\right)\left|I_{i}\right|$$

If $\lim _{|P| \rightarrow 0} \sigma(f, P, \xi)$exist and finite, we often denoted it as $\int_I f(x)\mathrm d x$ or :

$$\int_{I} f\left(x^{1}, \cdots, x^{n}\right) \mathrm{d} x^{1} \cdots \mathrm{d} x^{n}, \quad \int \cdots \int f\left(x^{1}, \cdots, x^{n}\right) \mathrm{d} x^{1} \cdots \mathrm{d} x^{n}$$

Fubini’s Theorem

Let $X \times Y$ be an interval in $\mathbb{R}^{m+n}$, which is the direct product of intervals $X \subset \mathbb{R}^{m}$ and $Y \subset \mathbb R^{n}$. If the function $f: X \times Y \rightarrow \mathbb{R}$ is integrable over $X \times Y$, then all three of the integrals

$$\int_{X \times Y} f(x, y) \mathrm{d} x \mathrm{d} y, \quad \int_{X} \mathrm{d} x \int_{Y} f(x, y) \mathrm{d} y, \quad \int_{Y} \mathrm{d} y \int_{X} f(x, y) \mathrm{d} x$$

exist and are equal.

Change of Variable in a Multiple Integral

If $\varphi: D_{t} \rightarrow D_{x}$ is a diffeomorphism of a bounded open set $D_{t} \subset \mathbb{R}^{n}$ onto a set $D_{x}=\varphi\left(D_{t}\right) \subset \mathbb{R}^{n}$ of the same type, $f \in \mathcal{R}\left(D_{x}\right),$ and $\operatorname{supp} f$ is a compact subset of $D_{x}$, then $f \circ \varphi\left|\operatorname{det} \varphi^{\prime}\right| \in \mathcal{R}\left(D_{t}\right)$, and the following formula holds:

simplified Chinese: 设 $\varphi: D_{t} \rightarrow D_{x}$ 为 $\mathbb{R}^{n}$ 中的有界开集之间的微分同胚，$f$ 为 定义在 $D_{x}$ 上的函数，满足条件 $\operatorname{supp} f \subset D{x} .$ 则 $f \in \mathcal{R}\left(D{x}\right)$ 当且仅当 $f \circ \varphi \cdot\left|\operatorname{det} \varphi^{\prime}\right| \in \mathcal{R}\left(D_{t}\right)$, 并且如果 $f \in \mathcal{R}\left(D_{x}\right)$，则成立如下的变量代换公式：

$$\int_{D_{x}} f(x) \mathrm{d} x=\int_{D_{t}} f \circ \varphi \cdot\left|\operatorname{det} \varphi^{\prime}\right| \mathrm{d} t$$

Example

(1) Polar coordinate transformation

In general polar coordinates $(r,\theta_1,\cdots,\theta_{n−1})$ in $\mathbb R^n$ are introduced via the relations:

$$\left\{\begin{array}{l} x^{1}\quad=r \cos \theta_{1} \\ x^{2}\quad=r \sin \theta_{1} \cos \theta_{2} \\ \qquad\vdots \\ x^{n-1}=r \sin \theta_{1} \sin \theta_{2} \cdot \ldots \sin \theta_{n-2} \cos \theta_{n-1} \\ x^{n}\quad=r \sin \theta_{1} \sin \theta_{2} \cdot \ldots \cdot \sin \theta_{n-1} \sin \theta_{n-1} \end{array}\right.$$

because of the orthogonality, we have:

$$\begin{aligned} \left|\operatorname{det} \psi^{\prime}\right|^{2} &=\operatorname{det} \psi^{\prime} \cdot \operatorname{det}\left(\psi^{\prime}\right)^{T}=\left\|\frac{\partial x}{\partial r}\right\|\left\|\frac{\partial x}{\partial \theta_{1}}\right\| \cdots\left\|\frac{\partial x}{\partial \theta_{n-1}}\right\| \\ &=\left(r^{n-1} \sin ^{n-2} \theta_{1} \sin ^{n-3} \theta_{2} \ldots \sin \theta_{n-2}\right)^{2} \end{aligned}$$

so we recall the Jacobian:

$$J=r^{n-1} \sin ^{n-2} \theta_{1} \sin ^{n-3} \theta_{2} \cdot \ldots \cdot \sin \theta_{n-2}$$

(2) The volume of $n-$dimensional sphere

Denote the sphere is $B_n(r)$We can easily have the recurrence formula:

$$V_{n}(r)=\int_{-r}^{r} \mathrm{d} x \int_{B_{n-1}\left(\sqrt{r^{2}-x^{2}}\right)} \mathrm{d} x^{2} \cdots \mathrm{d} x^{n}=\int_{-r}^{r} V_{n-1}\left(\sqrt{r^{2}-x^{2}}\right) \mathrm{d} x$$

It's trivial that $V_1(r)=2r$, $V_2(r)=\pi r^2$, assume that $V_{n}(r)=c_{n} r^{n}$, so

$$\begin{aligned} V_{n+1}(r) &=\int_{-r}^{r} c_{n}\left(r^{2}-x^{2}\right)^{\frac{n}{2}} \mathrm{d} x=c_{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^{n} \cos ^{n} \varphi \cdot r \cos \varphi \mathrm{d} \varphi \\ &=c_{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^{n+1} \cos ^{n+1} \varphi \mathrm{d} \varphi=c_{n+1} r^{n+1} \end{aligned}$$

with

$$c_{n+1}=c_{n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{n+1} \varphi \mathrm{d} \varphi$$

so that,

$$V_{n}(r)=\left\{\begin{array}{ll} 2 \frac{(2 \pi)^{k}}{(2 k+1) ! !} r^{2 k+1}, & n=2 k+1 \\ \frac{(2 \pi)^{k}}{(2 k) ! !} r^{2 k}, & n=2 k \end{array}\right.$$

$\blacksquare$